3.368 \(\int (a+a \sec (c+d x))^{3/2} (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=101 \[ \frac {8 a^2 (5 B+3 C) \tan (c+d x)}{15 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a (5 B+3 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{15 d}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d} \]

[Out]

2/5*C*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/d+8/15*a^2*(5*B+3*C)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/15*a*(5*B+3
*C)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d

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Rubi [A]  time = 0.13, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {4054, 12, 3793, 3792} \[ \frac {8 a^2 (5 B+3 C) \tan (c+d x)}{15 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a (5 B+3 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{15 d}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[c + d*x])^(3/2)*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(8*a^2*(5*B + 3*C)*Tan[c + d*x])/(15*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*(5*B + 3*C)*Sqrt[a + a*Sec[c + d*x]]*T
an[c + d*x])/(15*d) + (2*C*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*d)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3792

Int[csc[(e_.) + (f_.)*(x_)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(-2*b*Cot[e + f*x])/
(f*Sqrt[a + b*Csc[e + f*x]]), x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]

Rule 3793

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(b*Cot[e + f*x]*(a
+ b*Csc[e + f*x])^(m - 1))/(f*m), x] + Dist[(a*(2*m - 1))/m, Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m - 1), x
], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, 1/2] && IntegerQ[2*m]

Rule 4054

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) +
(a_))^(m_.), x_Symbol] :> -Simp[(C*Cot[e + f*x]*(a + b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[1/(b*(m + 1)),
Int[(a + b*Csc[e + f*x])^m*Simp[A*b*(m + 1) + (a*C*m + b*B*(m + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b,
e, f, A, B, C, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)]

Rubi steps

\begin {align*} \int (a+a \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac {2 C (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}+\frac {2 \int \frac {1}{2} a (5 B+3 C) \sec (c+d x) (a+a \sec (c+d x))^{3/2} \, dx}{5 a}\\ &=\frac {2 C (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}+\frac {1}{5} (5 B+3 C) \int \sec (c+d x) (a+a \sec (c+d x))^{3/2} \, dx\\ &=\frac {2 a (5 B+3 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{15 d}+\frac {2 C (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}+\frac {1}{15} (4 a (5 B+3 C)) \int \sec (c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {8 a^2 (5 B+3 C) \tan (c+d x)}{15 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a (5 B+3 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{15 d}+\frac {2 C (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}\\ \end {align*}

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Mathematica [A]  time = 0.29, size = 62, normalized size = 0.61 \[ \frac {2 a^2 \tan (c+d x) \left ((5 B+9 C) \sec (c+d x)+25 B+3 C \sec ^2(c+d x)+18 C\right )}{15 d \sqrt {a (\sec (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sec[c + d*x])^(3/2)*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(2*a^2*(25*B + 18*C + (5*B + 9*C)*Sec[c + d*x] + 3*C*Sec[c + d*x]^2)*Tan[c + d*x])/(15*d*Sqrt[a*(1 + Sec[c + d
*x])])

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fricas [A]  time = 0.45, size = 89, normalized size = 0.88 \[ \frac {2 \, {\left ({\left (25 \, B + 18 \, C\right )} a \cos \left (d x + c\right )^{2} + {\left (5 \, B + 9 \, C\right )} a \cos \left (d x + c\right ) + 3 \, C a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{15 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^(3/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

2/15*((25*B + 18*C)*a*cos(d*x + c)^2 + (5*B + 9*C)*a*cos(d*x + c) + 3*C*a)*sqrt((a*cos(d*x + c) + a)/cos(d*x +
c))*sin(d*x + c)/(d*cos(d*x + c)^3 + d*cos(d*x + c)^2)

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giac [A]  time = 2.91, size = 170, normalized size = 1.68 \[ \frac {4 \, {\left ({\left (2 \, \sqrt {2} {\left (5 \, B a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 3 \, C a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 5 \, \sqrt {2} {\left (5 \, B a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 3 \, C a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 15 \, \sqrt {2} {\left (B a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + C a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{15 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^(3/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

4/15*((2*sqrt(2)*(5*B*a^4*sgn(cos(d*x + c)) + 3*C*a^4*sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)^2 - 5*sqrt(2)*(5
*B*a^4*sgn(cos(d*x + c)) + 3*C*a^4*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 + 15*sqrt(2)*(B*a^4*sgn(cos(d*x
+ c)) + C*a^4*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - a)^2*sqrt(-a*tan(1/2*d*x +
1/2*c)^2 + a)*d)

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maple [A]  time = 1.54, size = 95, normalized size = 0.94 \[ -\frac {2 \left (-1+\cos \left (d x +c \right )\right ) \left (25 B \left (\cos ^{2}\left (d x +c \right )\right )+18 C \left (\cos ^{2}\left (d x +c \right )\right )+5 B \cos \left (d x +c \right )+9 C \cos \left (d x +c \right )+3 C \right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, a}{15 d \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))^(3/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

-2/15/d*(-1+cos(d*x+c))*(25*B*cos(d*x+c)^2+18*C*cos(d*x+c)^2+5*B*cos(d*x+c)+9*C*cos(d*x+c)+3*C)*(a*(1+cos(d*x+
c))/cos(d*x+c))^(1/2)/cos(d*x+c)^2/sin(d*x+c)*a

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^(3/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

Timed out

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mupad [B]  time = 7.58, size = 213, normalized size = 2.11 \[ -\frac {2\,a\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}-1\right )\,\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left (B\,25{}\mathrm {i}+C\,18{}\mathrm {i}+B\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,10{}\mathrm {i}+B\,{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,50{}\mathrm {i}+B\,{\mathrm {e}}^{c\,3{}\mathrm {i}+d\,x\,3{}\mathrm {i}}\,10{}\mathrm {i}+B\,{\mathrm {e}}^{c\,4{}\mathrm {i}+d\,x\,4{}\mathrm {i}}\,25{}\mathrm {i}+C\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,18{}\mathrm {i}+C\,{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,48{}\mathrm {i}+C\,{\mathrm {e}}^{c\,3{}\mathrm {i}+d\,x\,3{}\mathrm {i}}\,18{}\mathrm {i}+C\,{\mathrm {e}}^{c\,4{}\mathrm {i}+d\,x\,4{}\mathrm {i}}\,18{}\mathrm {i}\right )}{15\,d\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B/cos(c + d*x) + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^(3/2),x)

[Out]

-(2*a*(exp(c*1i + d*x*1i) - 1)*(a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(B*25i + C*18i +
B*exp(c*1i + d*x*1i)*10i + B*exp(c*2i + d*x*2i)*50i + B*exp(c*3i + d*x*3i)*10i + B*exp(c*4i + d*x*4i)*25i + C*
exp(c*1i + d*x*1i)*18i + C*exp(c*2i + d*x*2i)*48i + C*exp(c*3i + d*x*3i)*18i + C*exp(c*4i + d*x*4i)*18i))/(15*
d*(exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}} \left (B + C \sec {\left (c + d x \right )}\right ) \sec {\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))**(3/2)*(B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Integral((a*(sec(c + d*x) + 1))**(3/2)*(B + C*sec(c + d*x))*sec(c + d*x), x)

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